hdu oj 2602 Bone Collector（背包）

最新推荐文章于 2021-01-18 14:47:36 发布

原创最新推荐文章于 2021-01-18 14:47:36 发布 · 268 阅读

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本文介绍了一个经典的骨收集者背包问题，目标是在限定体积内获取最大价值的骨头。输入包括案例数、每种骨头的价值与体积，输出为能获得的最大总价值。

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 51860 Accepted Submission(s): 21837

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).

Sample Input

Sample Output

题目地址：点击打开链接

放松一下：陈奕迅《你的背包》

AC代码：

#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
int n, v, val[1005];
int cost[1005], f[1005];
int main()
{
	int t;scanf("%d", &t);
	while(t--)
	{
		int i, j;
		scanf("%d %d", &n, &v);
		memset(f, 0, sizeof(f));
		for(i = 1; i <= n; i++)
		    scanf("%d", &val[i]);
		for(i = 1; i <= n; i++)
		    scanf("%d", &cost[i]);
		for(i = 1; i <= n; i++)
		{
			for(j = v; j >= cost[i]; j--)
			{
				f[j] = max(f[j], f[j-cost[i]]+val[i]);
			}
		}
		printf("%d\n", f[v]);
	}
	return 0;
}

Sample Output